What method would you choose to solve the equation 2x2 – 7 = 9? Explain why you chose this method.

2 answers:

7 0

Normal solution;
By taking -7 to the RHS, you will have
$2 x^{2} =16$
dividing both sides by 2 gives;
$x^{2} =8
x= \sqrt{8} 
x=2 \sqrt{2}$

Kisachek [45]3 years ago

5 0

This is a quadratic equation . We order to equate to zero

2x² – 7 = 9

2x² - 7-9 = 0
2x² - 16 = 0

This equation is a binomial
2* (x² - 8) = 0
We choose the simplest method . The equation with practice it can be calculated in the memory

The product is equal to 0 if one of the factors is zero

2≠0 ∨ x² - 8 = 0
x² = √8
x = √8 ∨ x= - √8

Answer : x = √8 ∨ x = -√8

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The radii of two circles are in the ratio of 3 to 1. Find the area of the smaller circle if the area of the larger circle is 27

Naddik [55]

ANSWER
$3\pi sq.\: in.$

EXPLANATION

Let R be the radius of the bigger circle and r, be the radius of the smaller circle.

Then their ratio is given as,

$R:r=3:1$

We can rewrite it as fractions to get,

$\frac{R}{r} = \frac{3}{1}$

We make R the subject to get,

$R = 3r$

The area of the bigger circle can be found using the formula,

$Area=\pi {r}^{2}$

This implies that,

$Area=\pi ({3r})^{2}$

$Area=9\pi {r}^{2}$

But it was given in the question that, the area of the bigger circle is 27π.

$27\pi=9\pi {r}^{2}$

We divide through by 9π to get,

$3 = {r}^{2}$

This means that,
$r = \sqrt{3}$

The area of the smaller circle is therefore

$= \pi {( \sqrt{3}) }^{2}$

$= 3\pi$